An iron rod is placed parallel to magnetic field of intensity $2000 \mathrm{~A} / \mathrm{m}$. The magnetic flux through the rod is $6 \times 10^{-4} \mathrm{~Wb}$ and its cross-sectional area is $3 \mathrm{~cm}^{2}$. The magnetic permeability of the rod in $\mathrm{Wb} / \mathrm{A}-\mathrm{m}$ isA) $10^{-1}$B) $10^{-2}$C) $10^{-3}$D) $10^{-4}$

Home » An iron rod is placed parallel to magnetic field of intensity . The magnetic flux through the rod is and its cross-sectional area is . The magnetic permeability of the rod in isA) B) C) D)

An iron rod is placed parallel to magnetic field of intensity $2000 \mathrm{~A} / \mathrm{m}$ . The magnetic flux through the rod is $6 \times 10^{-4} \mathrm{~Wb}$ and its cross-sectional area is $3 \mathrm{~cm}^{2}$ . The magnetic permeability of the rod in $\mathrm{Wb} / \mathrm{A}-\mathrm{m}$ is
A) $10^{-1}$
B) $10^{-2}$
C) $10^{-3}$
D) $10^{-4}$

Physics

An iron rod is placed parallel to magnetic field of intensity $2000 \mathrm{~A} / \mathrm{m}$ . The magnetic flux through the rod is $6 \times 10^{-4} \mathrm{~Wb}$ and its cross-sectional area is $3 \mathrm{~cm}^{2}$ . The magnetic permeability of the rod in $\mathrm{Wb} / \mathrm{A}-\mathrm{m}$ is
A) $10^{-1}$
B) $10^{-2}$
C) $10^{-3}$
D) $10^{-4}$

Answer is (C) $\mu=\mathrm{B} \cdot \mathrm{H}=\frac{\phi}{\mathrm{A}} \times \mathrm{H}=\frac{6 \times 10^{-4}}{3 \times 10^{-4}} \times \frac{1}{2 \times 10^{3}}=10^{-3}$

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